292

ELECTRICAL APPARATUS

This differential equation is integrated by:

5 = Aeco,
which, substituted in (12) gives:

aAeCe + AC*tCe = 0,

168. 1. If a <0, it is:

where :

« + <72 = 0, __
C ± V~- a.

(13)

(14)

m

/ —
= V- a=

Since in this case, e+m° is continually increasing, the syn-
chronous motor is unstable. That is, without oscillation, the
synchronous motor drops out of step, if /3 > a.

2. If a > 0, it is, denoting:

or, substituting for e+jri6 and t+jnQ the trigonometric functions:

5 = (Ax + A 2) cos nB + j (Ai - A2) sin n^,
or,

= B cos

+ r)-

That is, the synchronous motor is in stable equilibrium, when
oscillating with a constant amplitude B, depending upon the
initial conditions of oscillation, and a period, which for small
oscillations gives the frequency of oscillation:

f _ „/ ~
/0 - nj -
As instance, let :
6Q = 2200 volts. Z = 1 +- 4 j ohms, or, » = 4.12; a = 76°.
And let the machine, a 16-polar, 60-cycle, 400-kw., revolving-
field, synchronous motor, have the radius of gyration of 20 in.,
a weight of the revolving part of 6000 Ib.
The momentum then is Jkf0 = 850,000 joules. •
Deriving the angles, /3, corresponding to given values of output,
P, and excitation, e, from the polar diagram, or from the symbolic